# CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 7 COORDINATE GEOMETRY EX – 7.4

## Coordinate Geometry

Page No: 171

1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).

Solution:

Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.

Coordinates of point of division can be given as follows:

x = and y =

Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have

2( + (– 4 = 0

+ ( = 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

Or k = 2/9

Hence, the ratio is 2:9.

2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution:

If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of a triangle = = 0[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0

• 2x – y + 7y – 14 = 0
• 2x + 6y – 14 = 0
• x + 3y – 7 = 0. Which is required result.

3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Solution:

Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA =

OB =

OC =

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly: OB = OC

(x – 3)+ (y + 7)2 = (x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y+ 14y + 49 = y– 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, centre of the circle located at point (3,-2).

4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution:

Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD

To Find: Coordinate of points B and D.

Step 1: Find distance between A and C and coordinates of point O.

We know that, diagonals of a square are equal and bisect each other.

AC = = 4

Coordinates of O can be calculated as follows:

x = (3-1)/2 = 1 and y = (2+2)/2 = 2

So O(1,2)

Step 2: Find the side of the square using Pythagoras theorem

Let a be the side of square and AC = 4

From right triangle, ACD,

a = 2√2

Hence, each side of square = 2√2

Step 3: Find coordinates of point D

Equate length measure of AD and CD

Say, if coordinate of D are

Squaring both sides,

Similarly, CD2 =

Since all sides of a square are equal, which means AD = CD

Value of y1 can be calculated as follows by using the value of x.

From step 2: each side of square = 2√2

CD2 =

8 =

Hence, D = (1, 4)

Step 4: Find coordinates of point B

From line segment, BOD

Coordinates of B can be calculated using coordinates of O; as follows:

Earlier, we had calculated O = (1, 2)

Say B =

For BD;

1 =

x= 1

And 2 =

=> y2 = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

Solution:

(i) Taking A as origin, coordinates of the vertices P, Q and R are,

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here AD is the x-axis and AB is the y-axis.

(ii) Taking C as origin,

Coordinates of vertices P, Q and R are ( 12, 2) , (13, 6) and (10, 3) respectively.

Here CB is the x-axis and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using formula: Area of a triangle =

= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ ( – 12 – 3 + 24 )

= 9/2 sq unit

(ii) Area of triangle PQR in case of origin C:

Area of a triangle =

= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= ½ ( 36 + 13 – 40)

= 9/2 sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

6. The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that . Calculate the area of the ∆ ADE and compare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Solution:

Given: The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2)

Point D and Point E divide AB and AC respectively in ratio 1 : 3.

Coordinates of D can be calculated as follows:

x = and y =

Here

Consider line segment AB which is divided by the point D at the ration 1:3.

x =

y =

Similarly, Coordinates of E can be calculated as follows:

x =

y = = 5

Find Area of triangle:

Using formula: Area of a triangle =

Area of triangle ∆ ABC can be calculated as follows:

= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

Area of ∆ ADE can be calculated as follows:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

Hence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16.

7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.

(iv) What do you observe?

[Note : The point which is common to all the three medians is called the centroid

and this point divides each median in the ratio 2 : 1.](v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Solution:

(i) Coordinates of D can be calculated as follows:

Coordinates of D =

D

(ii) Coordinates of P can be calculated as follows:

Coordinates of P =

P

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = = (5/2 , 3)

E(5/2 , 3)

Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:

Coordinates of Q =

F is the mid- point of the side AB

Coordinates of F =

Point R divides the side CF in ratio 2:1

Coordinates of R =

(iv) Coordinates of P, Q and R are same which shows that medians intersect each other at a common point, i.e. centroid of the triangle.

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:

8. ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Solution:

P id the mid-point of side AB,

Coordinate of P =

Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5,

Coordinate of S = (2, -1)

Now,

Length of PQ = = =

Length of SP = = =

Length of QR = = =

Length of RS = = =

Length of PR (diagonal) = = 6

Length of QS (diagonal) = = 5

The above values show that, PQ = SP = QR = RS = , i.e. all sides are equal.

But PR ≠ QS i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.